How do I convert an MDP with the reward function in the form $R(s,a,s')$ to and an MDP with a reward function in the form $R(s,a)$?
I think I may be in position to answer my own question. The Bellman equation (for the optimal policy) for a MDP with $r(s,a,s')$ rewards would look like this:

$$V(s) = \max_a \left\{ \sum_{s'} p(s'|s,a)(r(s,a,s') + \gamma V(s')) \right\} $$
$$V(s) = \max_a \left\{ \sum_{s'} p(s'|s,a) \cdot r(s,a,s') + \gamma \sum_{s'} p(s'|a,s) \cdot V(s') \right\} $$

Now, $ \sum_{s'} p(s'|s,a) \cdot r(s,a,s') $ is precisely $ \mathbb{E}\left[ r(s,a,s') | s,a \right] = r(s,a) $.

So all in all, the resulting Bellman equation looks like this:

$$V(s) = \max_a \left\{r(s,a) + \gamma \sum_{s'} p(s'|s,a) \cdot V(s') \right\} $$

It is clear, then, that a process with $ r(s,a,s') $ rewards can be transformed to a $ r(s,a) $ process without introducing artificial states and maintaining the optimal policies.

As a side note unrelated to the question itself, that leads me to believe that $ r(s,a,s') $ functions may be convenient in some scenarios, but they do not add "expressive power", in the sense that they do not allow to model problems more compactly (as it happens when we extend $ r(s) $ to $ r(s,a) $).